1. Relations and Functions
ALGEBRA 2 LESSON 2-1
(For help, go to Skills Handbook page 848 and Lesson 1-2.)
Graph each ordered pair on the coordinate plane.
1. (–4, –8) 2. (3, 6) 3. (0, 0) 4. (–1, 3) 5. (–6, 5)
Evaluate each expression for x = –1, 0, 2, and 5.
6. x –2x x |x – 3|
2-1

2. Relations and Functions
ALGEBRA 2 LESSON 2-1
Solutions 1. 2. 3. 4. 5.
2-1

3. Relations and Functions
ALGEBRA 2 LESSON 2-1
Solutions (continued)
6. x + 2 for x = –1, 0, 2, and 5:
–1 + 2 = 1; = 2; = 4; = 7
7. –2x + 3 for x = –1, 0, 2, and 5:
–2(–1) + 3 = = 5; –2(0) + 3 = = 3;
–2(2) + 3 = –4 + 3 = –1; –2(5) + 3 = – = –7
8. 2x2 + 1 for x = –1, 0, 2, and 5:
2 • (–1)2 + 1 = 2 • = = 3;
2 • = 2 • = = 1;
2 • = 2 • = = 9;
2 • = 2 • = = 51
9. |x – 3| for x = –1, 0, 2, and 5:
|–1 – 3| = |–4| = 4; |0 – 3| = |–3| = 3;
|2 – 3| = |–1| = 1; |5 – 3| = |2| = 2
2-1

4. Relations and Functions
ALGEBRA 2 LESSON 2-1
Graph the relation {(–3, 3), (2, 2), (–2, –2), (0, 4), (1, –2)}.
Graph and label each ordered pair.
2-1

5. Relations and Functions
ALGEBRA 2 LESSON 2-1
Write the ordered pairs for the relation. Find the domain and range.
{(–4, 4), (–3, –2), (–2, 4), (2, –4), (3, 2)}
The domain is {–4, –3, –2, 2, 3}.
The range is {–4, –2, 2, 4}.
2-1

6. Relations and Functions
ALGEBRA 2 LESSON 2-1
Make a mapping diagram for the relation {(–1, 7), (1, 3), (1, 7), (–1, 3)}.
Pair the domain elements with the range elements.
2-1

7. Relations and Functions
ALGEBRA 2 LESSON 2-1
Determine whether the relation is a function.
The element –3 of the domain is paired with both
4 and 5 of the range.
The relation is not a function.
2-1

8. Relations and Functions
ALGEBRA 2 LESSON 2-1
Use the vertical-line test to determine whether the graph represents a function.
If you move an edge of a ruler from left to right across the graph, keeping the edge vertical as
you do so, you see that the edge of the ruler never intersects the graph in more than one point in any position.
Therefore, the graph does represent a function.
2-1

9. Relations and Functions
ALGEBRA 2 LESSON 2-1
Find ƒ(2) for each function.
a. ƒ(x) = –x2 + 1
ƒ(2) = – = –4 + 1 = –3
b. ƒ(x) = |3x|
ƒ(2) = |3 • 2| = |6| = 6
c. ƒ(x) = 9
1 – x
ƒ(2) = = = –9 9
1 – 2 –1
2-1

10. Relations and Functions
ALGEBRA 2 LESSON 2-1
Pages 59–61 Exercises 1. 2. 3. 4.
5. (–2, –2), (–1, 1), (1, 1), (1, 0), (3, 3), (3, –2);
domain {–2, –1, 1, 3}, range {–2, 0, 1, 3}
6. (–2, 3), (0, 1), (2, –1), (3, –2); domain {–2, 0, 2, 3},
range {–2, –1, 1, 3}
7. (–2, 0), (–1, 2), (0, 3), (1, 2), (2, 0); domain {–2, –1, 0, 1, 2}, range {0, 2, 3} 8.
2-1

11. Relations and Functions
ALGEBRA 2 LESSON 2-1 9.
10.
11.
12. not a function
13. function
14. function
15. function
16. function
17. not a function
18. function
19. function
20. not a function
21. function
22. –7, –3, 4, 11
23. 13, 7, –3.5, –14
, 6.5, 10, 13.5
25. –2, –4, –7.5, –11
26. 21, 13, –1, –15
27. –13, –9, –2, 5
, 17 , –3 , –24
29. – , – , , 1 3 1 3 2 3 2 3 23 6 13 6 3 4 11 3
2-1

12. Relations and Functions
ALGEBRA 2 LESSON 2-1 9 2 7 2 7 4
30. – , – , – , 0
31. y = , where y
is the number of meters and x is the number of inches; m
32.
domain {2, 3, 4, 5}, range {4, 5, 6, 7}
33.
domain {–4, –3, –2, –1}, range {1, 2, 3, 4}
34.
domain { – , 0, 2},
range { –2, – , , 2}
35.
domain { – , , , }
range { – , }
36. domain {2, 4, 8}, range {4, 8, 16}; function
37. domain {–2, –1, 0, 9}, range {2, 5, 7}; not a function x
39.37 3 2 1 2 3 2 5 2 1 2 1 2 1 2 1 2 1 2
2-1

13. Relations and Functions
ALGEBRA 2 LESSON 2-1
38. domain {all real numbers}, range {y 0}; function
39. domain {– x }, range {–1 y 1}; not a function
40. A
41. B
42. C
43. yes
44. no
45. no
46. v(s) = s3; cm3
47. v(r) = r 3; about 4849 cm3
48. No; since 2 in the domain maps to 1 and 4, it is not a function.
49. Check students’ work.
50. –3
51. –
52. –4x – 14
53. 7
54.
55. Yes; each x is paired with a unique y.
56. No; each positive x is paired with two y values. 2 3
> –
< –
< –
< –
< – 3 4
2-1

14. Relations and Functions
ALGEBRA 2 LESSON 2-1
57. Yes; each x is paired with a unique y.
58. Function; if the domain and range are interchanged, it is not a function because 6 would be paired with both 2.5 and 3.0.
59. Domain {all integers}, range {all even integers}; function, each integer pairs to a unique even integer.
60. Domain {all integers}, range {all integers}; function, each integer pairs with its opposite.
61. Domain {all integers}, range {all even integers}; not a function, each nonzero integer pairs with a pos. and neg. even integer.
62. C
63. G
64. [2] ƒ(5); ƒ(5) = –2(5) + 3 = –7 and g(–2) =
4(–2) – 3 = –11, and –7 > –11.
[1] only includes answer ƒ(5) or g(–2)
65. [4] Yes; S(c) = 6c2; S(2.5) = 6(2.5)2 = ; the surface area is 37.5 cm2.
[3] minor computational error
[2] includes only function
[1] no work shown
2-1

15. Relations and Functions
ALGEBRA 2 LESSON 2-1
66.
67.
68.
69. 0
70.
71.
72. 1 11
73. –6, 14
74. –2 b 2
75. x 6
% increase
77. 50% increase
78. 25% decrease 2 11 4 11
< –
< – 6 11 7 11 = / 3 11
2-1

16. Relations and Functions
ALGEBRA 2 LESSON 2-1
1. Write the ordered pairs for the relation. Find the domain and range.
2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function.
3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.
2-1

17. Relations and Functions
ALGEBRA 2 LESSON 2-1
4. Determine whether the graph represents a function.
5. Find ƒ(–5) for each function.
a. ƒ(x) = 5x + 35
b. ƒ(x) = x2 – x
2-1

18. Relations and Functions
ALGEBRA 2 LESSON 2-1
1. Write the ordered pairs for the relation. Find the domain and range.
2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function.
3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.
{(–3, 2), (–2, 0), (–1, –1), (2, 1), (2, 3), (4, 1)};
D = {–3, –2, –1, 2, 4], R = {–1, 0, 1, 2, 3}
function
(–4, 2) or (–4, 6)
2-1

19. Relations and Functions
ALGEBRA 2 LESSON 2-1
4. Determine whether the graph represents a function.
5. Find ƒ(–5) for each function.
a. ƒ(x) = 5x + 35
b. ƒ(x) = x2 – x
not a function 10 30
2-1

20. Evaluate each expression for x = –2, 0, 1, and 4. 1. x + 7 2. x – 2
Linear Equations
ALGEBRA 2 LESSON 2-2
(For help, go to Lesson 1-2.)
Evaluate each expression for x = –2, 0, 1, and 4.
x x – 2
3. 3x x – 8 2 3 3 5 1 2
2-2

21. Linear Equations Solutions 1. x + 7 for x = –2, 0, 1, 4:
ALGEBRA 2 LESSON 2-2
Solutions
1. x + 7 for x = –2, 0, 1, 4:
(–2) + 7 = – = = or 5 ;
(0) + 7 = = 7;
(1) + 7 = = 7 or ;
(4) + 7 = = = or 9 2 3 2 3 4 3 21 3
–4 + 21 3 17 3 2 3 2 3 2 3 2 3 2 3 23 3 2 3 8 3 21 3
8 + 21 3 29 3 2 3
2-2

22. Solutions (continued) 2. x – 2 for x = –2, 0, 1, 4:
Linear Equations
ALGEBRA 2 LESSON 2-2
Solutions (continued)
x – 2 for x = –2, 0, 1, 4:
(–2) – 2 = – + – = = – or –3 ;
(0) – 2 = 0 – 2 = –2;
(1) – 2 = + – = = – = – or –1 ;
(4) – 2 = – = = 3 5 3 5 6 5 10 5
–6 + (–10) 5 16 5 1 5 3 5 3 5 3 5 10 5
3 + (–10) 5 7 5 7 5 2 5 3 5 12 5 10 5
12 + (–10) 5 2 5
3. 3x + 1 for x = –2, 0, 1, 4:
3(–2) + 1 = –6 + 1 = –5; 3(0) + 1 = = 1;
3(1) + 1 = = 4; 3(4) + 1 = = 13
2-2

23. Solutions (continued) 4. x – 8 for x = –2, 0, 1, 4:
Linear Equations
ALGEBRA 2 LESSON 2-2
Solutions (continued)
4. x – 8 for x = –2, 0, 1, 4:
(–2) – 8 = –1 – 8 = –9; (0) – 8 = 0 – 8 = –8;
(1) – 8 = – 8 = –7 or – ;
(4) – 8 = 2 – 8 = –6 1 2 1 2 1 2 1 2 1 2 1 2 15 2 1 2
2-2

24. Graph the equation y = – x + 2.
Linear Equations
ALGEBRA 2 LESSON 2-2
Graph the equation y = – x + 2. 4 3
If x = 0, then y = 2.
and then draw the line through these two points.
Plot the points (0, 2) and (3, –2)
If x = 3, then y = –2.
2-2

25. Set x or y equal to zero to find each intercept.
Linear Equations
ALGEBRA 2 LESSON 2-2
The equation 10x + 5y = 40 models how you can give $.40 change if you have only dimes and nickels. The variable x is the number of dimes, and y is the number of nickels. Graph the equation. Describe the domain and the range. Explain what the x- and y-intercepts represent.
10x + 5y = 40 10x + 5y = 40
10x + 5(0) = 40 10(0) + 5y = 40
10x = 40 5y = 40
x = 4 y = 8
Set x or y equal to zero to find each intercept.
2-2

26. Use the intercepts to graph the equation.
Linear Equations
ALGEBRA 2 LESSON 2-2
(continued)
Use the intercepts to graph the equation.
The x-intercept is (4, 0), which means that the change can be given using 4 dimes and 0 nickels.
The possible solutions for this situation are limited to those points on the line segment connecting (0, 8) and (4, 0) whose x- and y-coordinates are whole numbers.
The y-intercept is (0, 8), which means that the change can be given using 0 dimes and 8 nickels.
The number of dimes and the number of nickels must each be a whole number.
Therefore, the domain is {0, 1, 2, 3, 4} and the range is {0, 2, 4, 6, 8}.
2-2

27. Find the slope of the line through the points (–2, 7) and (8, –6).
Linear Equations
ALGEBRA 2 LESSON 2-2
Find the slope of the line through the points (–2, 7) and (8, –6).
Slope = Use the slope formula.
y2 –y1
x2 – x1
=    Substitute (–2, 7) for (x1, y1) and (8, –6) for (x2, y2).
–6 – 7
8 – (–2)
= – Simplify. 13 10
The slope of the line is – 13 10
2-2

28. y – y1 = m(x – x1) Use the point-slope equation.
Linear Equations
ALGEBRA 2 LESSON 2-2
Write in standard form an equation of the line with slope 3 through the point (–1, 5).
y – y1 = m(x – x1) Use the point-slope equation.
y – 5 = 3[x – (–1)] Substitute 3 for m, 5 for y1, and –1 for x1.
y – 5 = 3[(x + 1)] Simplify.
y – 5 = 3x + 3 Distributive Property
3x – y = –8 Write in standard form.
2-2

29. y – y1 = m(x – x1) Write the point-slope equation.
Linear Equations
ALGEBRA 2 LESSON 2-2
Write in point-slope form an equation of the line through (4, –3) and (5, –1).
y – y1 = m(x – x1) Write the point-slope equation.
y – y1 = (x – x1) Substitute the slope formula for m.
y2 – y1
x2 – x1
y – (–3) = (x – 4) Substitute: x1 = 4, y1 = –3, x2 = 5, y2 = –1.
–1 – (–3)
5 – 4
y + 3 = 2(x – 4) Simplify.
You can also use (5, –1) for (x1, y1) and (4, –3) for (x2, y2).
This gives the equation y + 1 = 2(x – 5). Both equations define the same line.
2-2

30. –7x + 2y = 8 Add 7x to both sides.
Linear Equations
ALGEBRA 2 LESSON 2-2
Find the slope of –7x + 2y = 8.
–7x + 2y = 8 Add 7x to both sides.
y = x + 4 Write in the slope-intercept form. 7 2
The slope of the line is . 7 2
2-2

31. m = – Find the negative reciprocal of 4.
Linear Equations
ALGEBRA 2 LESSON 2-2
Write an equation of the line through (5, –3) and perpendicular to y = 4x + 1. Graph both lines.
m = – Find the negative reciprocal of 4. 1 4
y = mx + b Use slope-intercept form.
y = – x + b Slope is – . 1 4 1 4
–3 = – (5) + b Substitute (5, –3) for (x, y).
–3 = – + b Simplify. 5
– = b Solve for b. 7 4
y = – x – Write the equation. 1 4 7
2-2

32. Linear Equations Pages 67–71 Exercises 1. 2. 3. 4. 5. 6. 2-2
ALGEBRA 2 LESSON 2-2
Pages 67–71 Exercises 1. 2. 3. 4. 5. 6.
2-2

33. c. 0.23 represents a cost of $.23 per mile driven.
Linear Equations
ALGEBRA 2 LESSON 2-2 7. 8.
9. a.
y = 0.23x
domain {x | x 0}
range {y | y 0}
10. a.
b. The x-intercept is the point that represents selling caps and 0 sweatshirts in order to raise the $4500. The y-intercept represents selling 0 caps and 360 sweatshirts to raise the $4500.
> –
> –
b. x-intercept (0, 0), y-intercept (0, 0); when no miles have been driven, there is no cost.
c represents a cost of $.23 per mile driven.
2-2

34. Linear Equations 11. –1 12. –2 13. 3 14. 15. – 16. 1 17. undefined
ALGEBRA 2 LESSON 2-2 4 3
11. –1
12. –2
13. 3
14.
15. –
16. 1
17. undefined
18. 0
19.
20. 3x – y = –2
x – y =
x + y = –
23. y = –2
24. x + y = 2
25. 5x – y = –2
26. y – 3 = –1(x + 10)
27. y – 0 = (x – 1)
28. y – 10 = – (x + 4)
29. y + 1 = – (x – 0)
30. y – 11 = 1(x – 7)
31. y – 9 = – (x – 1)
32. –5
33.
34. –
35. –
36.
37. 0 5 6 19 3 3 5 12 5 7 5 4 11 1 5 3 2 1 2 A B 5 4 A B 17 5 5 2
2-2

35. Linear Equations 38. y = –3x – 5 39. y = x + 40. y = 10 41. x = 1 42.
ALGEBRA 2 LESSON 2-2
38. y = –3x – 5
39. y = x + 5 2 13
40. y = 10
41. x = 1
42.
43.
44.
2-2

36. 58. undefined slope, no y-intercept, (–3, 0)
Linear Equations
ALGEBRA 2 LESSON 2-2 1 4
45.
46.
47.
48.
49.
50.
51.
52.
53. –
54. , (0, 4), (–6, 0)
55. –1, (0, 1000), (1000, 0)
, 0, – , , 0
57. 5, (0, –1), , 0
58. undefined slope, no y-intercept, (–3, 0)
59. 0, (0, 0), all pts. on
x-axis 2 3 R S T S T R 1 5 1 3 2 3
2-2

37. 69. a. I, II; III; graphs I and II show constant rate of change.
Linear Equations
ALGEBRA 2 LESSON 2-2 1 2 5 2 3 2 1 2
60. – , 0, – , (–5, 0)
61. –0.8, (0, 0.4), (0.5, 0)
62. – , 0, , , 0
63. –
64.
65. –
66. y = x + 3
67. y = 3x + 2
68. y = – x –
69. a. I, II; III; graphs I and II show constant rate of change.
b. I and III
c. II
70. Vertical lines cannot be graphed by this method.
71. y = –1 
72. y = 2x + 1 
73. y = x +   A B C B C A 5 6 10 3 5 13 7 5 7 10 3 4
2-2

38. 78. a–e. Check students’ work. The polygon is a rectangle. 79. yes
Linear Equations
ALGEBRA 2 LESSON 2-2 3 2
74. y = – x – 1
75. 3x – 2y = 2
76. 9x + 3y = 2
77. 3x + 12y = –4
78. a–e. Check students’ work.
The polygon is a rectangle.
79. yes
80. no
81. a.
b. y = 3x + 6
c. y = – x +
d. They are perpendicular.
82. The equation of the line connecting (1, 3) to (–2, 6) is y = –x + 4. The equation of the line connecting (1, 3) to (3, 5) is y = x + 2. The slopes are negative reciprocals so the lines are perpendicular. Therefore by def. of a right triangle it is a right triangle. 1 3 8 3
2-2

39. 83. The slope of the line connecting (2, 5) to (4, 8) is , (2, 5) to
Linear Equations
ALGEBRA 2 LESSON 2-2
83. The slope of the line connecting (2, 5) to
(4, 8) is , (2, 5) to
(5, 3) is – , (4, 8) to
(7, 6) is – , and
(5, 3) to (7, 6) is .
Since the adjacent sides’ slopes are negative reciprocals they are perpendicular. By the def. of a rectangle, it is a rectangle.
84. p: y = 4x + 16
q: y = – x +
r : y = 4
85. A
86. G
87. C
88. B
89. C
90. B
91. domain {–2, 1, 2, 3, 4}, range {–2, –1, 2, 3}; not a function
92. domain {all reals}, range {all reals}; function
93. domain {–3, 0, 1, 7}, range {–10, –5, –1, 3}; not a function 1 4 13 4 3 2 2 3 2 3 3 2
2-2

40. 94. Commutative Prop. of Add.
Linear Equations
ALGEBRA 2 LESSON 2-2
94. Commutative Prop. of Add.
95. multiplicative inverses
96. Distributive Prop.
97. additive inverses, additive identity
98. studio: $6.26; designer: $9.39
2-2

41. 1. Find the slope of the line through the points (–5, –1) and (2, 3).
Linear Equations
ALGEBRA 2 LESSON 2-2
1. Find the slope of the line through the points (–5, –1) and (2, 3).
2. Write an equation in standard form for the line with slope 3 through (9, –4).
3. Write in point-slope form an equation of the line through the points (–3, 8) and (7, 6). Use (–3, 8) as the point for the equation.
4. Write the equation 3x – 12y = 6 in slope-intercept form.
5. What is the slope of a line perpendicular to y = x – 7? What is the slope of a line parallel to y = x – 7? 4 7
3x – y = 31
y – 8 = – (x + 3) 1 5
y = x – 1 4 2 2 3 2 3 3 2 ; –
2-2

42. Solve each equation for y. 1. 12y = 3x 2. 12y = 5x 3. y = 15
Direct Variation
ALGEBRA 2 LESSON 2-3
(For help, go to Lesson 1-3 and Skills Handbook page 844.)
Solve each equation for y.
1. 12y = 3x 2. 12y = 5x 3. y = 15
4. 0.9y = 27x 5. 5y = 35 3 4
Tell whether each equation is true. 1 4 2 8 5 12 36 20 24 30 6 15 9
2-3

43. Direct Variation Solutions 1. 12y = 3x y = = = 2. 12y = 5x y = = x
ALGEBRA 2 LESSON 2-3
Solutions
1. 12y = 3x
y = = = 3x 12
3 • x
3 • 4 x 4
2. 12y = 5x
y = = x 5x 12 5
y = 15
y = = = 20 3 4 60
4. 0.9y = 27x
y = = 30x
27x
0.9
5. 5y = 35
y = 7 6.
1(8) 4(2)
8 = 8
true 1 4 2 8
2-3

44. Solutions (continued)
Direct Variation
ALGEBRA 2 LESSON 2-3
Solutions (continued) 7.
2(15) 5(6)
30 = 30
true 2 5 6 15 8.
9(36) 24(12)
false 12 36 9 24 9.
20(36) 24(30)
720 = 720 20 30 = /
2-3

45. = and are both equal to –3, but = 3.
Direct Variation
ALGEBRA 2 LESSON 2-3
For each function, determine whether y varies directly with x. If so, find the constant of variation and write the equation.
x –
y 3 –6 15 a.
= and are both equal to –3, but = 3. y x 3 –1 –6 2 15 5
Since the three ratios are not all equal, y does not vary directly with x.
x –4
y –8 b.
= 2, so y does vary directly with x. y x 14 7 18 9 –8 –4 =
The constant of variation is 2.
The equation is y = 2x.
2-3

46. 5x = –2y is equivalent to y = – x, so y varies directly with x.
Direct Variation
ALGEBRA 2 LESSON 2-3
For each function, tell whether y varies directly with x. If so, find the constant of variation.
a. 3y = 7x + 7
Since you cannot write the equation in the form y = kx, y does not vary directly with x.
b. 5x = –2y
5x = –2y is equivalent to y = – x, so y varies directly with x. 5 2
The constant of variation is – . 5 2
2-3

47. a. Find the constant of variation.
Direct Variation
ALGEBRA 2 LESSON 2-3
The perimeter of a square varies directly as the length of a side of the square. The formula P = 4s relates the perimeter to the length of a side.
a. Find the constant of variation.
The equation P = 4s has the form of a direct variation equation with k = 4.
b. Find how long a side of the square must be for the perimeter to be 64 cm.
P = 4s Use the direct variation.
64 = 4s Substitute 64 for P.
16 = s Solve for s.
The sides of the square must have length 16 cm.
2-3

48. 15(18) = 27(y) Write the cross products.
Direct Variation
ALGEBRA 2 LESSON 2-3
Suppose y varies directly with x, and y = 15 when x = 27. Find y when x = 18. Let (x1, y1) = (27, 15) and let (x2, y2) = (18, y).
Write a proportion. y1 x1 y2 x2 =
Substitute. = 15 27 y 18
15(18) = 27(y) Write the cross products.
y = Solve for y.
15 • 18 27
y = 10 Simplify.
2-3

49. Direct Variation Pages 74–77 Exercises 1. yes; k = 2, y = 2x
ALGEBRA 2 LESSON 2-3
Pages 74–77 Exercises 2 7 10 7
1. yes; k = 2, y = 2x
2. yes; k = –3, y = –3x
3. no
4. yes; k = , y = x
5. yes; k = 7, y = 7x
6. no
7. yes; k = –2, y = –2x
8. no
9. yes; k = 12
10. yes; k = 6
11. yes; k = –2
12. no
13. no
14. yes; k = –5
15. yes; k = 6
16. no
17. k = ; –
18. k = – ;
19. k = –1; 5
20. k = 2; –10
21. k = – ; 21
22. k = – ; 1
23. a. k =
b. s = h
c ft 1 in. 5 3 25 3 1 3 1 3 17 4 1 4 1 4 1 4 13 36 13 36
2-3

50. 46. No; y = 1.7x does not contain the point (9, –9).
Direct Variation
ALGEBRA 2 LESSON 2-3
24. –3
25. 4
27.
mi/h
29. yes; k = , y = x
30. no
31. no
32. yes; k = 1.3, y = 1.3x
33. y = 2x
34. y = x
35. y = – x
36. y = –500x
37. y = x
38. y = – x
39. y = x
40. y = – x
41. 9
42. 6
43. 90
44. –140
46. No; y = 1.7x does not contain the point (9, –9).
47. Yes; y = – x contains the point (15, –12 ).
48. Yes; y = x contains the point (6 , 22 ). 7 3 9 2 5 3 3 5 2 3 2 3 1 9 5 6 2 7 1 2 14 3 7 2 1 2 3 4
2-3

51. 49. Answers may vary. Sample: y = 0.5x
Direct Variation
ALGEBRA 2 LESSON 2-3
49. Answers may vary. Sample: y = 0.5x
50. Answers may vary. Sample: y = 3.2x
51. Answers may vary. Sample:  y = – x
52. a. y = 28x
b mi
c gal
d. $0.056/mi
53. Answers may vary. Sample: No; y = 0 passes through the origin, but is not a direct variation.
54. Answers may vary. Sample: If y varies directly with x2, and y = 2 when x = 4,
then y = when
x = 9.
55. y is doubled.
56. y is halved.
57. y is divided by 7.
58. y is multiplied by 10.
59. a.
b. 32
c. z = kxy, x = k1w, so z = kk1wy, z varies jointly with w and y.
60. B 1 2 3 4 1 8
2-3

52. 64. [2] No; the ratio is not constant.
Direct Variation
ALGEBRA 2 LESSON 2-3
65.
66.
67.
61. G
62. D
63. I
64. [2] No; the ratio is not constant.
[1] correct answer with no explanation
68.
69.
70.
domain {4, 7},
range {0, –1} y x
2-3

53. Direct Variation 71. domain {1, 2, 4, 5}, range {–2, –1, 1, 2} 72.
ALGEBRA 2 LESSON 2-3
71.
domain {1, 2, 4, 5},
range {–2, –1, 1, 2}
72.
domain {1, 2, 3, 4},
range {7, 8, 9, 10} %
2-3

54. Direct Variation
ALGEBRA 2 LESSON 2-3
1. For each function, tell whether y varies directly as x. If so, find the constant of variation and write the equation.
a.  b.
2. Determine whether y varies directly as x. If so, find the constant of variation.
a. y = 7x + 4
b. y = x
c. 2y = –12x
x y
–4 16 –
8 –32
x y
2 10
3 15
6 25 5 3
2-3

55. Direct Variation
ALGEBRA 2 LESSON 2-3
3. Assume y varies directly as x. If y = 8 when x = 42, find y when x = 126.
4. How can you tell from the graph of a linear function whether the function is a direct variation?
2-3

56. Direct Variation
ALGEBRA 2 LESSON 2-3
1. For each function, tell whether y varies directly as x. If so, find the constant of variation and write the equation.
a.  b.
2. Determine whether y varies directly as x. If so, find the constant of variation.
a. y = 7x + 4
b. y = x
c. 2y = –12x
x y
–4 16 –
8 –32
x y
2 10
3 15
6 25
yes; –4; y = –4x no no 5 3
yes; 5 3
yes; –6
2-3

57. Direct Variation
ALGEBRA 2 LESSON 2-3
3. Assume y varies directly as x. If y = 8 when x = 42, find y when x = 126.
4. How can you tell from the graph of a linear function whether the function is a direct variation? 24
If the graph is a non-horizontal line through the origin, then the linear function is a direct variation.
2-3

58. Find the change in x and the change in y between each pair of points.
Using Linear Models
ALGEBRA 2 LESSON 2-4
(For help, go to Lessons 2-1 and 2-2.)
Find the change in x and the change in y between each pair of points.
1. (–0.2, 9) and (3.4, 7.3) 2. (10, 17) and (11.5, 13.5)
3. (0, ) and (–1, )
4. ƒ(x) = x – 2 for x = –3, 0,
5. g(x) = 3(2 – x) for x = 0, , 1 3 10 2 5
Evaluate each function for the given values. 4 3 1 2 1 6
2-4

59. Using Linear Models Solutions 1. (– 0.2, 9) and (3.4, 7.3)
ALGEBRA 2 LESSON 2-4
1. (– 0.2, 9) and (3.4, 7.3)
change in x: 3.4 – (–0.2) = = 3.6
change in y: 7.3 – 9 = –1.7
2. (10, 17) and (11.5, 13.5)
change in x: 11.5 – 10 = 1.5
change in y: 13.5 – 17 = –3.5
3. 0, and –1,
change in x: –1 – 0 = –1
change in y: – = – = 3 10 2 5 4 1
Solutions
2-4

60. Solutions (continued)
Using Linear Models
ALGEBRA 2 LESSON 2-4
4. ƒ(x) = x – 2 for x = –3, 0, :
ƒ(–3) = (–3) – 2 = –4 – 2 = –6;
ƒ(0) = (0) – 2 = 0 – 2 = –2;
ƒ = – 2 = – = = – = – or –1
5. g(x) = 3(2 – x) for x = 0, , 1:
g(0) = 3(2 – 0) = 3(2) = 6;
g = 3 2 – = = = = = or 5 ;
g(1) = 3(2 – 1) = 3(1) = 3 4 3 1 2 6 12
4 – 12 8 –
12 – 1 11 33
Solutions (continued)
2-4

61. Relate: plane’s elevation = rate • time + starting elevation .
Using Linear Models
ALGEBRA 2 LESSON 2-4
Suppose an airplane descends at a rate of 300 ft/min from an elevation of 8000 ft.
Write and graph an equation to model the plane’s elevation as a function of the time it has been descending. Interpret the intercept at which the graph intersects the vertical axis.
Relate: plane’s elevation = rate • time + starting elevation .
Define: Let t = time (in minutes) since the plane began its descent.
Let d = the plane’s elevation.
Write: d = –300 • t
2-4

62. An equation that models the plane’s elevation is d = –300t + 8000.
Using Linear Models
ALGEBRA 2 LESSON 2-4
(continued)
An equation that models the plane’s elevation is d = –300t
The d-intercept is (0, 8000).
This tells you that the elevation of the plane was 8000 ft at the moment it began its descent.
2-4

63. Step 1: Identify the two points as (x1, y1) and (x2, y2).
Using Linear Models
ALGEBRA 2 LESSON 2-4
A spring has a length of 8 cm when a 20-g mass is hanging at the bottom end. Each additional gram stretches the spring another 0.15 cm. Write an equation for the length y of the spring as a function of the mass x of the attached weight. Graph the equation. Interpret the y-intercept.
Step 1: Identify the two points as (x1, y1) and (x2, y2).
Adding another 20 g of mass at the end of the spring will give a total mass of 40 g and a length of (20) = 11 cm. Use the points (x1, y1) = (20, 8) and (x2, y2) = (40, 11) to find the linear equation.
2-4

64. Step 2: Find the slope of the line. m = Use the slope formula.
Using Linear Models
ALGEBRA 2 LESSON 2-4
(continued)
Step 2: Find the slope of the line.
m = Use the slope formula.
m = Substitute.
m = , or 0.15 Simplify.
y2 – y1
x2 – x1
11 – 8
40 – 20 3 20
Step 3: Use one of the points and the point-slope form to write an equation for the line.
y – y1 = m(x – x1) Use the point-slope form.
y – 8 = 0.15(x – 20) Substitute.
   y = 0.15x + 5 Solve for y.
2-4

65. Using Linear Models (continued)
ALGEBRA 2 LESSON 2-4
(continued)
An equation of the line that models the length of the spring is y = 0.15x + 5.
The y-intercept is (0, 5). So, when no weight is attached to the spring, the length of the spring is 5 cm.
2-4

66. y = 0.15x + 5 Write the equation.
Using Linear Models
ALGEBRA 2 LESSON 2-4
Use the equation from Additional Example 2. What mass would be needed to stretch the spring to a length of 9.5 cm?
y = 0.15x + 5 Write the equation.
9.5 = 0.15x + 5 Substitute 9.5 for y.
= x Solve for x.
9.5 – 5
0.15
30 = x Simplify.
The mass should be 30 g.
2-4

67. A linear model seems reasonable since the points fall close to a line.
Using Linear Models
ALGEBRA 2 LESSON 2-4
An art expert visited a gallery and jotted down her guesses for the selling price of five different paintings. Then, she checked the actual prices. The data points (guess, actual) show the results, where each number is in thousands of dollars. Graph the data points. Decide whether a linear model is reasonable. If so, draw a trend line and write its equation.
{(12, 11), (7, 8.5), (10, 12), (5, 3.8), (9, 10)}
A linear model seems reasonable since the points fall close to a line.
Trend lines and equations may vary.
2-4

68. Using Linear Models (continued)
ALGEBRA 2 LESSON 2-4
(continued)
A possible trend line is the line through (6, 6) and (10.5, 11). Using these two points to write an equation in slope-intercept form gives y = x – . 10 9 2 3
2-4

69. Using Linear Models Pages 81–84 Exercises 1. d = 62.5h + 15
ALGEBRA 2 LESSON 2-4
Pages 81–84 Exercises
1. d = 62.5h + 15
2. a. y = –50t b.
The y-intercept (0, 1000) represents the filled pool, and the t-intercept (20, 0) represents the time needed to empty the pool.
3. h = 8x + 60
4. y = x – 1; 5 leaves
5. y = 0.5x ; lb
6. y = 58.3x – 3.3; 172 blades of grass
2-4

70. Linear model is reasonable; models may vary. Sample: y = –1.3x + 11 9.
Using Linear Models
ALGEBRA 2 LESSON 2-4
7. y = 1.75x ; $8.40 8.
Linear model is reasonable; models may vary. Sample: y = –1.3x + 11 9.
Linear model is reasonable; models may vary. Sample:
y = 2.6x – 0.6
10.
Linear model is reasonable; models may vary. Sample: y = –0.75x – 3.7
11.
not reasonable
2-4

71. 12. a. Linear model is reasonable.
Using Linear Models
ALGEBRA 2 LESSON 2-4
12. a.  Linear model is reasonable.
b. 40
c. Answers may vary. Sample: After drawing a trend line on the graph, locate the European size on the y-axis. Then find the corresponding U.S. size on the x-axis.
13. a.
Answers may vary. Sample:
y = 125x + 975
b cal
c. Answers may vary. Sample: No; adults need fewer Calories, not more.
2-4

72. b–e. Answers may vary. Samples are given. b. c = 0.04A
Using Linear Models
ALGEBRA 2 LESSON 2-4
14. a.
b–e. Answers may vary. Samples are given.
b. c = 0.04A
c. The model fits the data very closely.
d. No; the area of the tarp is 150 ft2 so the price should be $6.00.
e. 6  8 ft; $0.07
15. y = –4x + 10
16. y = –3x – 6
17. y = –7.5x – 2.5
18. y = x –
2-4

73. b. y = 2.95x; slope = 2.95, y-intercept = 0
Using Linear Models
ALGEBRA 2 LESSON 2-4
19. a. y = 29.95
b. y = 2.95x; slope = 2.95, y-intercept = 0
c. Answers may vary. Sample:
Either way, you will average the same costs over the long run.
20. a. Answers may vary. Sample: y = x – 9.3
b g
c. 200 Cal; a 200 Cal hamburger is closer to 5 g.
2-4

74. Using Linear Models 21. a. population 22. 104.5 b–c. 23. 85.8 24. 6.5
ALGEBRA 2 LESSON 2-4
21. a. population
b–c. 
d. 2 million
e. Answers may vary. Sample: Strong; the points fall close to a straight line.
25. 13
2-4

75. c. Answers may vary. Sample: neither; y = –1.5x + 12
Using Linear Models
ALGEBRA 2 LESSON 2-4 17 11
27. a.
b. about $3900
c. about $1400
d. Answers may vary. Sample: No, expenditure would be predicted to be about $5000.
28. a. y = – x
b. y = –2x + 10
c. Answers may vary. Sample: neither; y = –1.5x + 12
29. y = x – 2 5 2
2-4

76. 35. a. Answers may vary. Sample: y = x + b. 70 36. –2.7; 13.5
Using Linear Models
ALGEBRA 2 LESSON 2-4 16 3
34. a. y = – x
b tons
35. a. Answers may vary. Sample:
y = x +
b. 70
36. –2.7; 13.5
37. –3; 15
38. – ; 6
; –42
40. {–7, –3, –1, 2, 7}
41. {1, 3.5, 5, 6, 8}
42. {3, 4, 5.25, 12, 19}
43. { }
44. {–299, –99, 1, 151, 401}
45. {1, 6, 9, 11, 15}
46. a ft/s; 2091 ft
b. d = 20.91t
c mi/h
– , – , –2, – , 0 7 2 5 4 16 3 2 3 6 5 42 5
2-4

77. b. In what year will the house be only 250 ft from the water?
Using Linear Models
ALGEBRA 2 LESSON 2-4
1. A family built a house in a beach resort area. In 1990, the house was 430 ft from the water. With erosion, the house was 400 ft from the water in 2000.
a.  Assuming a linear relationship between the number of years x since 1990 and the distance y of the house from the water, write an equation for y as a function of x.
b. In what year will the house be only 250 ft from the water?
2. Graph the set of data. Decide whether a linear model is reasonable. If so, draw a trend line and write its equation.
{(1, 2), (3, 3), (3, 3.75), (4, 4), (5, 3.25), (6, 4.5)}
y = –3x + 430
2050
A linear model seems reasonable;
Answers may vary.
Sample: y = 0.4x +2
2-4

78. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
(For help, go to Lesson 2-2 and Skills Handbook page 851.)
Graph each equation for the given domain and range.
1. y = x for real numbers x and y 0
2. y = 2x – 4 for real numbers x and y 0
3. y = –x + 6 for real numbers x and y 3
> –
> –
< –
2-5

79. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
Solutions
1. y = x                    
x = real numbers
y 0
2. y = 2x – 4              
3. y = –x + 6             
y 3
> –
> –
< –
2-5

80. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
Graph y = |2x – 1| by using a table of values.
Evaluate the equation for several values of x.
x –1.5 –1 – y
Make a table of values.
Graph the function.
2-5

81. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
Graph y = |x – 1| – 1 on a graphing calculator.
Use the absolute value key. Graph the equation
Y1 = abs(X – 1) – 1
2-5

82. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
Use the definition of absolute value to graph y = |3x + 6| – 2.
Step 1: Isolate the absolute value.
  y = |3x + 6| – 2
y + 2 = |3x + 6|
Step 2: Use the definition of absolute value. Write one equation for 3x and a second equation for 3x + 6 < 0.
> –
when 3x
y + 2 = 3x + 6
y = 3x + 4
> –
when 3x + 6 < 0
y + 2 = –(3x + 6)
y = –3x – 8
2-5

83. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
(continued)
Step 3: Graph each equation for the appropriate domain.
When 3x , or x –2, y = 3x + 4.
When 3x + 6 < 0, or x < –2, y = –3x – 8.
> –
> –
2-5

84. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
A train traveling on a straight track at 50 mi/h passes a certain crossing halfway through its journey each day. Sketch a graph of its trip based on its distance and time from the crossing.
The equation d = |50t| models the train’s distance from the crossing.
2-5

85. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
Pages 88–90 Exercises
1–9. Tables may vary. Samples are given. 1. 2. 3. 4. 5.
2-5

86. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5 6. 7. 8. 9.
10.
11.
2-5

87. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
12.
13.
14.
15.
16.
17.
18.
19.
20.
2-5

88. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
21.
22.
23.
24.
25.
26.
27.
2-5

89. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
28.
29. B
30. C
31. A
32. D
33.
34.
35.
36.
37.
38.
2-5

90. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
39.
40.
41.
42.
43.
44.
45.
46.
47.
2-5

91. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
48.
49.
50.
51. a. time and distance before and after the roadside stand
b. A
52. a.
b. Answers may vary. Sample: Same shape and size, different vertices,
and one points down.
53.
54.
2-5

92. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
55.
56.
57.
58.
59. Answers may vary. Samples:
a. y = |x|, y = –|x|
b. y = |x|, y = |x – 1| + 1
60. B
61. I
62. A
63. I
64. C
65. [2] The vertex of y = |3x – 6| would be where 3x – 6 = –(3x – 6) because that would be where the graphs of both lines meet. In this case it is at x = 2.
[1] only includes solution x = 2
2-5

93. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
66.
[4] y = –5x – 1, y = 5x + 1
[3] does not label vertex
[2] does not label vertex and does not include both equations
[1] y = –|5x + 1| is not interpreted correctly but graph is correct
67.
Answers may vary.
Sample: y = 8.7x – 10.9
2-5

94. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
68.
Answers may vary.
Sample: y = x + 6
69.
not reasonable
70.
Answers may vary.
Sample: y = –0.5x + 7
71. –3
72. 4
73. –
74. 4 3 2
75. –
76. 5
77. y = 18h; linear
78. 11
79. 9
80. 13
81. 4
82. 7 2 3 2 3 1 9
2-5

95. Absolute Value Functions and Graphs
ALGEBRA 2 LESSON 2-5
1. Graph y = –|x + 3|.
2. Graph y = |2x – 6| – 2.
2-5

96. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
(For help, go to Lessons 2-2 and 2-5.)
1. y = x, y = x y = –2x, y = –2x – 3
3. y = |x|, y = |x| – 2 4. y = –|2x|, y = –|2x| ± 1
5. y = |x|, y = |x – 1| 6. y = – x , y = – x + 2
Graph each pair of functions on the same coordinate plane. 1 2 1 2
2-6

97. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Solutions
1. y = x
y = x + 4
3. y = |x|
y = |x| – 2
5. y = |x|
y = |x – 1|
2. y = –2x
y = –2x – 3
4. y = – |2x|
y = – |2x| + 1
6. y = – x
y = – x + 2 1 2 1 2
2-6

98. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Graph y = |2x| and y = |2x| + 3 on the same coordinate plane. Describe how the graphs are related. Make a table of values and graph the equation.
x y = |2x| y = |2x| + 3 – –
For each value of x, y = |2x| + 3 is 3 more than the value of y = |2x|.
The graph of y = |2x| + 3 is the graph of y = |2x| shifted 3 units up.
2-6

99. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Identify the parent function for y = 2x – 3 and the value of k. Then, graph the function by translating the parent function.
The parent is y = 2x, and k = –3.
Translate the graph of y = 2x down 3 units.
2-6

100. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Write an equation to translate the graph of y = |4x|, 5 units down.
The graph of y = |4x|, shifted 5 units down, means k = –5.
An equation is y = |4x| – 5.
2-6

101. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Identify the parent function and the value of h for y = | x – 5|. Graph both functions.
The parent function is y = |x|, and h = 5.
The minus sign means to move to the right. Translate the graph of y = |x| right 5 units.
2-6

102. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
The graph is a translation of y = –|x|. Write an equation for the graph.
The given graph can be obtained by translating the graph of y = –|x|, 3 units left.
So an equation for the given graph is y = –| x + 3|.
2-6

103. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Describe a possible translation of Figures M and N in the design shown below.
Translate Figure M 2 units down.
For Figure N, there are two possible translations: 4 units down, or 1 unit right and 2 units down.
2-6

104. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Graph y = | x + 1| + 2.
The parent function is y = |x|, h = 1, and k = 2.
The first plus sign means move 1 unit to the left.
The second plus sign means move k units up.
Place the vertex at (–1, 2) and draw the graph opening upward.
2-6

105. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Write an equation for each translation of y = |x|.
a. 3 units up, 7 units right
7 units right h = 7; minus sign
3 units up k = 3; plus sign
An equation is y = | x – 7| + 3.
b. 5 units down, 1 unit left
1 unit left h = 1; plus sign
5 units down k = 5; minus sign
An equation is y = | x + 1| – 5.
2-6

106. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
Pages 95–98 Exercises
1. y = –|x| + 3 is y = –|x| 3 units up.
2. f(x) = |x| – 1 is f(x) = |x| one unit down.
3. g(x) = |3x| + 2 is g(x) = |3x| 2 units up. is
units down.
5. y = x, k = 3
6. y = –x, k = 4
7. y = |x|, k = 2
8. y = –|x|, k = 6 1 2 2 3
y = |x| –
y = |x| 1 2 3
2-6

107. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6 2 3
9. y = x –
10. y = |x| + 4
11. y = –|x| + 2
12. y = |x|, h = 4
13. y = |x|, h = 5
14. y = –|x|, h = 1
15. y = –|x|, h = 2
16. y = |x + 2|
17. y = |x – 3|
18. y = –|x – 4|
19. Answers may vary. Sample: 1 unit left and 1 unit up
20. Answers may vary.
Sample: units left
and 1 unit up 6 5
2-6

108. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6 5 2 1 2
21.
22.
23.
24.
25.
26.
27. y = |x – 3| – 1
28. y = –|x + 1| + 2
29. y = – |x – | +
30. y = |x – 7| + 3
31. vertical
32. diagonal
2-6

109. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
33. horizontal
y = |x + 1|
34. vertical
35. vertical
36. vertical
37. diagonal
38. diagonal
39. diagonal
40. y = |x| – 3
41. y = x + 2
42. y = |x – 3| – 2
43. h = 1, k = 0
44. h = 0, k = 3
2-6

110. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
45. h = 2, k = 2
46. Horizontal; the graph is shifted left.
47. Answers may vary. Sample: Since a vert. translation affects only y-values and a horiz. translation affects only x-values, order is irrelevant.
48. a.
b. y = 3x + 5
c. y = 3x – 1
d. E
e. y = m(x – h) + k
49. Check students’ work.
50. y = |x – 7| – 2
51. y = –|x| – 3
52. y = –|x + 5|
53. y = x + 7
54. y = |x – a| + b
55. y = x + (q – p)
56.
6 right
2-6

111. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
57.
4 right, 3 down
58.
3 left, 1 up
59.
3 right, 3 up
60. a.
b. Answers may vary. Sample: It is a function because each t is paired with a unique height.
c. vertical translation 5 ft up
d. C
61. a. b. c.
62. C
63. H
64. A
2-6

112. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
65. [4] The graph of y = |x + 3| – 2 is y = |x| first translated 3 units to the left for y = |x + 3| and then 2 units down for y = |x + 3| – 2.
[3] includes equations and graph with no explanation
[2] includes either the translation equations OR graph
[1] attempts to interpret the translation for
y = |x + 3| – 2 and graphs it but makes an error and does not translate it 3 units left and 2 units down
66. Answers may vary. Sample: ƒ(0) = 5; ƒ(1) = 4; ƒ(2) = 3; ƒ(3) = 2; ƒ(–1) = 6
2-6

113. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
67. Answers may vary. Sample: ƒ(0) = –2;
ƒ(1) = 0; ƒ(2) = 2; ƒ(–2) = 0; ƒ(–1) = –2
y = |2x + 1| – 3
68. Answers may vary. Samples: ƒ(0) = 1;
ƒ(1) = ; ƒ(3) = ;
ƒ(–1) = ; ƒ(–3) =
69. x –10
70. a > 4.5
71. b > –1.5
• 75 • x = 281,250; 25 ft 8 9 2 3 10 9 4 3
< –
2-6

114. Vertical and Horizontal Translations
ALGEBRA 2 LESSON 2-6
1. Graph y = |x| and y = |x + 4| + 3 on the same coordinate plane. Describe the translation that takes the graph of the parent function to the graph of the other function.
Translate the parent graph 4 units left and 3 units up.
2. Write equations for the graphs obtained by translating y = |x| and y = –|x| as described.
a.  10 units right
b.  4 units down
c.  7 units left, 6 units up
y = |x – 10|; y = –|x – 10|
y = |x| – 4; y = –|x| – 4
y = |x + 7| + 6; y = –|x + 7| +6
2-6

115. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
(For help, go to Lessons 1-4 and 1-5.)
Solve each inequality. Graph the solution on a number line.
1. 12p t > – 2t 11
4. |4c| = |5 – b| = 3 6. |2h| 7
< –
> –
Solve and graph each absolute value equation or inequality.
> –
2-7

116. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
Solutions
1. 12p 15 p p
3. 5 – 2t 11
–2t 6
t –3
5. | 5 – b | = 3
5 – b = 3 or 5 – b = –3
–b = –2 or –b = – 8
b = 2 or b = 8
< –
t > 17
t > 13
4. |4c| = 18
4c = 18 or 4c = –18
c = 4.5 or   c = –4.5
6. |2h| 7
2h or 2h –7
h or h –
h or h –3.5 15 12
< –
< –
> –
> –
< –
> –
> –
< – 7 2 7 2
> –
< –
> –
< –
2-7

117. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7 3 2
Graph y > x + 1.
Step 1: Graph the boundary line y = x + 1. Since the inequality is greater than, not greater than or equal to, use a dashed boundary line. 3 2
Step 2: Since the inequality is greater than, y-values must be more than those on the boundary line. Shade the region above the boundary line.
2-7

118. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
A restaurant has only 15 eggs until more are delivered. An order of scrambled eggs requires 2 eggs. An omelet requires 3 eggs. Write an inequality to model all possible combinations of orders of scrambled eggs and omelets the restaurant can fill till more eggs arrive. Graph the inequality.
Relate:
plus
is less than or equal to
number of eggs needed for x orders of scrambled eggs
number of eggs needed for y orders of omelets 15
Define: Let x = the number of orders for scrambled eggs.
Let y = the number of orders for omelets.
Write: x + 3 y 15
< –
2-7

119. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
(continued)
Step 1: Find two points on the boundary line. Use the points to graph the boundary line.
when y = 0, 2x + 3(0) = 15
2x = 15
x = 15 2
when x = 0, 2(0) + 3y = 15
3y = 15
y = 5
Graph the points ( , 0) and (0, 5). Since the inequality is less than or equal to, use a solid boundary line. 15 2
2-7

120. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
(continued)
Step 2: Since the inequality is less than, y-values must be less than those on the boundary line. Shade the region below the boundary line.
All ordered pairs with whole-number coordinates in the shaded area and on the boundary line represent a combination of x orders of scrambled eggs and y orders of omelets that the restaurant could fill.
2-7

121. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
Graph y |2x| – 3.
> –
Since the inequality is greater than or equal to, the boundary is solid and the shaded region is above the boundary.
2-7

122. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
Write an inequality for each graph. The boundary line is given.
a. Boundary: y = | x – 2| – 1
The boundary line is dashed. The shaded region is above the boundary. This is the graph of y > |x – 2| – 1.
b. Boundary: y = – x + 3 1 2
The boundary is solid. The shaded region is below the boundary. This is the graph of y – x + 3. 1 2
< –
2-7

123. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
Pages 102–104 Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9.
2-7

124. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
10. a. y x if x 6,
y x if x > 6
b. Answers may vary. Sample:
> –
< –
11.
12.
13.
14.
15.
16.
> –
2-7

125. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
17.
18.
19.
20. y < –x – 2
21. 5x + 3y 9
22. 2y |2x + 6|
23.
24.
y > x + 2
25.
26.
27.
< –
> – 2 5
2-7

126. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
28.
29.
30.
31.
32.
33.
34.
35.
2-7

127. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
36.
37.
38. x > –3
39. y x + 2
40. y –2x + 4
41. y |x + 2|
42. y < –|x – 4|
43. y > |x + 1| – 1
44. Answers may vary. Sample:
y – x +
> –
45. a. c + s
b. c + s c.
Yes, (20, 50) is a solution.
d. 140 sundaes
46. Answers may vary. Sample: when it lies on the boundary line
< –
< – 5 3 95 3
< –
< – 3 2
2-7

128. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
47.
48.
49.
50.
51. A
52. I
53. D
54. F
55. C
2-7

129. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
56. [2] y x, where y is the number of tornadoes that could occur in the next x years. The domain is all whole numbers greater than 0, since years are whole numbers and can’t be negative. The range is whole numbers , since you cannot have a fraction of a tornado and in the first year you will have at least 300.
[1] includes only inequality with no explanation of domain and range
57.
> –
58.
59.
60.
61.
> –
2-7

130. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
62.
63. no
64. yes; 100
65. yes; –5
66. no
67. yes; 3
68. no
69. yes; –10
70. no
71. $7800
72.
73.
74.
2-7

131. Two-Variable Inequalities
ALGEBRA 2 LESSON 2-7
Graph each inequality.
1. y – x + 3
> – 1 4
2. –y + 1 > – | x + 3 |
2-7

132. Functions, Equations and Graphs
ALGEBRA 2 CHAPTER 2 1 2
1. domain {0, 1, 2, 3, 4}, range {–16, –9, –4, –1, 0}
2. domain {3, 4, 5, 6, 7},
range {2, 3, 4, 5, 6}
3. domain {–2, 1, 3, 4 }
range {1, 2, 3, 4, 5}
4. domain {– , 0.3, 5}
range {–0.01, 0, 2.5}
5. 1
6. 4
7. 1
8. 12
9. –17 1 2
2-A

133. Functions, Equations and Graphs
ALGEBRA 2 CHAPTER 2
10. 22
11. Check students’ work.
12. 2
13. –
14.
15. 3x + y = 0
16. 2x – 5y = –23
17. 4x – y = –3
18. x + 2y = 12
19. Answers may vary. Sample:
y = – x
20. Answers may vary. Sample: y + 6 = –16(x + 1)
21. Answers may vary. Sample:
y = (x – 3)
22. Answers may vary. Sample: y – 2 = 3(x – 8)
23. Answers may vary. Sample:
a. y = –x
b. y = x + 15
c. y = x
d. y = –x – 5
e. rectangle
24. 8; –4
; –
min 7 4 4 3 1 2 1 2 2 3 1 3
2-A

134. Functions, Equations and Graphs
ALGEBRA 2 CHAPTER 2
27.
28.
29.
30.
31. a–b.
Answers may vary.
Sample y = 222x ,
where 1990 corr. to x = 0
  c. Answers may vary. Sample: $4011
d. Check students’ work.
2-A

135. Functions, Equations and Graphs
ALGEBRA 2 CHAPTER 2
32. vertical
33. diagonal
34. diagonal
35. vertical
36. vertical
37. diagonal
38.
39.
2-A

136. Functions, Equations and Graphs
ALGEBRA 2 CHAPTER 2
40.
41.
2-A